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Chapter 14: Problem 21

It takes \(880 \mathrm{J}\) to raise the temperature of \(350 \mathrm{g}\) of lead from 0 to \(20.0^{\circ} \mathrm{C} .\) What is the specific heat of lead?

Short Answer

Expert verified
Answer: The specific heat of lead is \(0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\).

Step by step solution

01

Identify the given values

We are given the following information: - Amount of energy required to raise the temperature (Q) = 880 J - Mass of lead (m) = 350 g - Initial temperature (T1) = 0°C - Final temperature (T2) = 20.0°C
02

Write down the formula for energy change

The formula for energy change, which relates energy, mass, specific heat, and temperature change is: \(Q = mc (T_{2} - T_{1})\) We need to solve for the specific heat of lead (c), so we can rewrite the formula as follows: \(c = \frac{Q}{m(T_{2} - T_{1})}\)
03

Plug in the given values into the formula

Now we can substitute the given values into the equation: \(c = \frac{880 \mathrm{J}}{350 \mathrm{g}(20.0^{\circ}\mathrm{C}-0^{\circ}\mathrm{C})}\)
04

Calculate the specific heat

Simplify and compute the value for the specific heat: \(c = \frac{880 \mathrm{J}}{350 \mathrm{g}(20.0^{\circ}\mathrm{C})}\) \(c = \frac{880 \mathrm{J}}{7000 \mathrm{g}\cdot^{\circ}\mathrm{C}}\) \(c = 0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\)
05

Write the final answer

The specific heat of lead is \(0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\).

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Most popular questions from this chapter

If \(125.6 \mathrm{kJ}\) of heat are supplied to \(5.00 \times 10^{2} \mathrm{g}\) of water at \(22^{\circ} \mathrm{C},\) what is the final temperature of the water?
A spring of force constant \(k=8.4 \times 10^{3} \mathrm{N} / \mathrm{m}\) is compressed by \(0.10 \mathrm{m} .\) It is placed into a vessel containing $1.0 \mathrm{kg}$ of water and then released. Assuming all the energy from the spring goes into heating the water, find the change in temperature of the water.
What mass of water at \(25.0^{\circ} \mathrm{C}\) added to a Styrofoam cup containing two 50.0 -g ice cubes from a freezer at \(-15.0^{\circ} \mathrm{C}\) will result in a final temperature of \(5.0^{\circ} \mathrm{C}\) for the drink?
A heating coil inside an electric kettle delivers \(2.1 \mathrm{kW}\) of electric power to the water in the kettle. How long will it take to raise the temperature of \(0.50 \mathrm{kg}\) of water from \(20.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) (tutorial: heating)
A 2.0 -kg block of copper at \(100.0^{\circ} \mathrm{C}\) is placed into $1.0 \mathrm{kg}$ of water in a 2.0 -kg iron pot. The water and the iron pot are at \(25.0^{\circ} \mathrm{C}\) just before the copper block is placed into the pot. What is the final temperature of the water, assuming negligible heat flow to the environment?
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